$1+i^2+i^4+i^6+\ldots+i^{2 n}$ is

JEE MATHEMATICS MCQ: COMPLEX NUMBERS

1 Easy

1 of 1

Understanding +4 | -1

1+i^2+i^4+i^6+\ldots+i^{2 n}

View answer and explanation

Correct option: D

Can't be determined

Option D (Can't be determined) is correct because the sum depends on the value of n, specifically whether n is even or odd.

Practice All Chapter MCQs →

{\left| {1 - i} \right|^x} = {2^x}

|z|=1

z \neq \pm 1

z, \frac{z}{1-z^2}

\left|z_r-r\right| \leq r, \forall r=1,2,3, \ldots, n

\left|\sum\limits_{r=1}^n z_r\right|

(z+a b)^3=a^3

a \neq 0

\alpha, \beta

\gamma

|\alpha-\beta|

{z^2} + {\left| z \right|^2} = 0,

z \in C

2x = - 1 + \sqrt 3 i

{\left( {1 - {x^2} + x} \right)^6} - {\left( {1 - x + {x^2}} \right)^6}